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Math Help - Continuous Branches of log z

  1. #1
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    Continuous Branches of log z

    Hey everyone.

    I am trying to define a continuous branch of log z in the
    complex plane slit along the positive axis: C\{[0,i*infinity]}

    I understand that we need to make a branch cut because log z will not be continuous on all of C. When the cut is the negative real axis I think I understand. And I also realize that the the cut can be any ray by my intuition. But I don't understand how to put it down as a function in this case.
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  2. #2
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    Quote Originally Posted by robeuler View Post
    Hey everyone.

    I am trying to define a continuous branch of log z in the
    complex plane slit along the positive axis: C\{[0,i*infinity]}

    I understand that we need to make a branch cut because log z will not be continuous on all of C. When the cut is the negative real axis I think I understand. And I also realize that the the cut can be any ray by my intuition. But I don't understand how to put it down as a function in this case.
    Define \log z = \ln |z| + i \arg (z) where \theta = \arg z is defined angle on -\tfrac{3\pi}{2} < \theta \leq \tfrac{\pi}{2}
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Define \log z = \ln |z| + i \arg (z) where \theta = \arg z is defined angle on -\tfrac{3\pi}{2} < \theta \leq \tfrac{\pi}{2}
    I see, it is more straightforward than I thought. Thanks a lot.

    How about this one:
    I have to show that
    |cosh z|^2 = cos^2(y)+sinh^2(x)

    I have tried writing out cosh z in terms of e^x and e^y so that I can take the modulus the old fashioned way (x^2+y^2)^(1/2).
    (from the e-definition of cosh z and using that e^z=e^xcos(y)+ie^xsin(y))

    but the resulting mess is overwhelming and I can't seem to split up the real and imaginary parts neatly.
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  4. #4
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    Quote Originally Posted by robeuler View Post
    I see, it is more straightforward than I thought. Thanks a lot.

    How about this one:
    I have to show that
    |cosh z|^2 = cos^2(y)+sinh^2(x)

    I have tried writing out cosh z in terms of e^x and e^y so that I can take the modulus the old fashioned way (x^2+y^2)^(1/2).
    (from the e-definition of cosh z and using that e^z=e^xcos(y)+ie^xsin(y))

    but the resulting mess is overwhelming and I can't seem to split up the real and imaginary parts neatly.
    If I did not mess up the identity it should be,
    \cosh (x+iy) = \cosh (x) \cosh (iy) + \sinh (x) \sinh (iy) = \cosh x \cos y + i\sinh x \sin y
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    If I did not mess up the identity it should be,
    \cosh (x+iy) = \cosh (x) \cosh (iy) + \sinh (x) \sinh (iy) = \cosh x \cos y + i\sinh x \sin y
    The identity is correct.
    so |coshz|^2= cosh^2x * cos^2y + sinh^2x * sin^2y.

    how do I get rid of the cosh and the sin terms?

    Edit: Solved: substitute 1-cos^2y for sin^2y and it follows pretty easily.
    Last edited by robeuler; October 6th 2008 at 09:59 PM.
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  6. #6
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    Here is one more I hope you might get to:

    If f(z) is an analytic branch of cosh^(-1)(z), find f'(z)

    Can I simply take the derivative of
    cosh^-1(z)= ln(x+(x-1)^(1/2)*(x+1)^(1/2))?

    in the end this equals 1/((x-1)^(1/2)*(x+1)^(1/2))
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